Theorem: Let $\mathbf{A}$ and $\mathbf{B}$ be categories, $F,G : \mathbf{A} \rightarrow \mathbf{B}$ functors and $\alpha : F \rightarrow G$ a natural transformation.

$\alpha$ is a natural isomorphism if and only if $\alpha_A : FA \rightarrow GA$ is an isomorphism (in $\mathbf{B}$) for each $A \in \mathbf{A}$.

Proof: Suppose $\alpha$ is a natural isomorphism. That is to say, it is an isomorphism between $F$ and $G$ in the category $[\mathbf{A}, \mathbf{B}]$. Then, we have a natural transformation $\beta : G \rightarrow F$ so that $\beta \alpha = 1_F$ and $\alpha \beta = 1_G$ (the identity natural transformations on $F$ and $G$, respectively). Recall $\alpha$ is a collection of arrows $\alpha_A : FA \rightarrow GA$ in $\mathbf{B}$ for each $A \in \mathbf{A}$. And, by definition, they compose "component-wise", so that $(\beta \alpha)_A = \beta_A \circ \alpha_A$ for each $A \in \mathbf{A}$. Then, clearly we have $(1_F)_A = \beta_A \circ \alpha_A$, but $(1_F)_A$ is, by definition, precisely the identity on $FA$ -- that is, $1_{FA}$. Simultaneously, we have $1_{GA} = (1_G)_A = \alpha_A \circ \beta_A$. These identities directly state that $\alpha_A$ (and $\beta_A$) are isomorphisms between $FA$ and $GA$.

Suppose that, for each $A \in \mathbf{A}$, $\alpha_A : FA \rightarrow GA$ is an isomorphism. Then, also for each $A$ we have an arrow $\beta_A : GA \rightarrow FA$ with $\beta_A \circ \alpha_A = 1_{FA}$ and $\alpha_A \circ \beta_A = 1_{GA}$. The identity natural transformation $1_F$ is defined so that $(1_F)_A = 1_{FA}$ for each $A$. So we have $(1_F)_A = \beta_A \circ \alpha_A$ for each $A$, which means that $\beta \circ \alpha = 1_F$. Similarly, we obtain $\alpha \circ \beta = 1_G$, and $\alpha$ (and $\beta$) are natural isomorphisms.